Basics of Network Theory (Part-I)

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Cornelius Derek Patrick
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1 Basics of Network Theory (PartI). A square waveform as shown in figure is applied across mh ideal inductor. The current through the inductor is a. wave of peak amplitude. V t (m sec) [Gate 987: Marks] V t m sec The current through the inductor is i L = vdt. The integration of a square wave L is a triangular wave so the current through the inductor is a triangular wave of volt peak amplitude. Slope of triangular wave is ± t 0. Two H inductance coils are connected in series and are also magnetically coupled to each other the coefficient of coupling being 0.. The total inductance of the combination can be (a) 0.4 H (b) 3. H M L L H H (c) 4.0 H (d) 4.4 H (d)

2 The equivalent inductance Leq = L L ± M = ± 0. M = K L L = 4 ± 0. 4 = 4. 4, The current i4 in the circuit of Figure is equal to i = 5A i = 3A I i 0 = 7A (a) A (b) A i 4 =? i 3 = 4A (c) 4 A (d) None of these [Gate 997: Mark] (b) I = i 0 i = A i 4 = A 4. The Voltage V in Figure is equal to

3 4V 5V 4V V (a) 3 V (b) 3 V (c) 5 V (d) None of these [Gate 997: Mark] (a) Apply KVL V 5 4 = 4 V = = 3V 5. The voltage V in Figure is always equal to A V 5V (a) 9 V (b) 5 V (c) V (d) None of these [Gate 997: Mark]

4 (d) V = V A 5 = V A 9 Since the voltage of A current source is not known, it is not possible to find the value of voltage V. 6. The voltage V in Figure is a 3 0V 5V b (a) 0 V (b) 5 V (c) 5 V (d) None of the these [Gate 997: Mark] (a) 7. In the circuit shown in the figure the current id through the ideal diode (zero cut in voltage and zero forward resistance) equals 4 i D 0V 4 A (a) 0 A (b) 4 A (c) A (d) None of these [Gate 997: 3 Marks]

5 (c) Applying the source conversion 5V i D V i D = 5 3 = Amp 8. The voltage across the terminals a and b in Figure is a V 3A b (a) 0.5 V (b) 3.0 V (c) 3.5 V (d) 4.0 V [Gate 998: Mark] (c) Taking b as reference node and applying KCL at a V ab V ab = 3 V ab V ab = 6 V ab = 6 = 3. 5V

6 9. In the circuit of the figure, the voltage V(t) is e at V(t) H e bt (a) e at e bt (b) e at e bt (c) ae at be bt (d) ae at be bt [Gate 000: Mark] (d) The voltage v(t) is v(t) = L d dt (eat e bt ) = ae at be bt 0. For the circuit in the figure, the voltage V0 is 4V V 0 V (a) V (b) V (c) V (d) None of these [Gate 000: Marks]

7 (d) Since the diode is forward biased, it is taken as short circuit. Writing KCL V 0 4 ( V 0 3V 0 = 0 ) ( V 0 ) = 0 V 0 = 3. The voltage e0 in the figure is 4 V 4 e 0 (a) V (b) 4/3 V (c) 4 V (d) 8 V [Gate 00: Marks] (c) Writing KCL e 0 4 3e 0 = e 0 = 4V e 0 4 e 0 4 = 0

8 . The voltage e0 in the figure is 8A 6V 0 6 e 0 (a) 48 V (b) 4 V (c) 36 V (d) 8 V [Gate 00: Mark] (d) Applying the source conversion, the circuit is as shown 0 6V 80V e 0 6 Writing KCL e 0 80 e e 0 = e 0 = 4 = 8V e 0 = 0 3. The dependent current source shown in the figure 5 V = 0V I 5 V 5 A

9 (a) Delivers 80W (b) absorbs 80 W (c) delivers 40 W (d) absorbs 40 W [Gate 00: Mark] (a) Writing KVL V 5I 5 (I V 5 ) = 0 V 0I V = 0 I = 0 Voltage across dependent source =0V Thus power delivered by it is = = 80W 4. The equivalent inductance measured between the terminals and for the circuit shown in the figure M L L (a) L L M (b) L L M (c) L L M (d) L L M [Gate 004: Mark] (d) The coils are wound in opposite directions, they are series opposing Leq = L L M

10 5. Impedance Z as shown in the figure is j5 j L j0 L j0 L 3 j Z (a) j 9 Ω (b) j 9 Ω (c) j 9 Ω (d) j 39 Ω [Gate 005: Marks] (b) Current leaves the dotted terminal of L and enters the dotted terminal of L3 so mutual induction M3 is negative, M3 is positive. jwleq = j5 j j j0 j0 = j9 6. If R = R = R4 = R and R3 =.R in the bridge circuit shown in the figure, then the reading in the ideal voltmeter connected between a and b is 0V a R R R 4 R 3 b (a) 0.38 V (b) 0.38 V (c) 0.38 V (d) V [Gate 005: Marks]

11 (c) V a = 0.R R R = 5V V b = 0.R 3 = 0. = R 4 R 3. V a V b = 0. 38V 7. In the circuit shown, the power supply by the voltage source is A A 0V (a) 0 W (b) 5 W (c) 0 W (d) 00 W [Gate 00: Marks] (a) Applying KVL in the outer loop (3I) I (I) A 3A A 0V

12 (3 I) ( I) = 0 6 I 4 I = 0 0 4I = 0 I = 0 Power supplied by the voltage source P = VI = 0 8. In the circuit shown below, the current I is equal to V I AC j4 j (a) A (b) A (c) A (d) A [Gate 0: Marks] (b) Converting delta into star the circuit is redrawn as impedance of the circuit is I V I AC j4 j4 ( j4) ( j4) = 7Ω Current I = = 0 0 A

13 9. The average power delivered to an impedance (4 j3)ω by a current 5 cos (00πt 00) A is (a) 44. W (c) 6.5 W (b) 50 W (d) 5 W [Gate 0: Mark] (b) Average power is same as rms power P = I rms R = ( 5 ) 4 = 50W 0. In the circuit shown below, the current through the inductor is V A i I j A D AC 0 0 V 0 0 V AC B j A (a) j A (b) j A (c) j A (d) 0 A C [Gate 0: Mark] (C) In the balanced bridge, the product of opposite arms are equal. j j = =

14 In the balanced bridge, current flowing through the diagonal element is zero. Applying nodal analysis at top node V 0 0 V 0 0 j = V j j V 0 0 = j V = j I = V 0 0 j = j j = j (j)j = j. Consider a delta connection of resistors and its equivalent star connection as shown below. If all elements of the delta connection are scaled by a factor k, k> 0, the elements of the corresponding star equivalent will be scaled by a factor of R a R C R B R b R c R A (a) K (b) K (c) /K (d) K [Gate 03: Mark] (b) In the star connection, R c = R A = R br a R a R b R c, R B = R ar c R a R b R c R b.r c R a R b R c If the delta connection components are scaled by a factor K, then the star equivalent will also be scaled by a factor K

15 . Find the current I in the following branch 5 5 0V A 0 I [Gate:04] Using superposition taking 0V source and replacing the current source by its internal resistance ( ) I = 0 = 0. 5A 0 Taking current source and replacing voltage source by its internal resistance zero I = 5 0 = 4 A Total current I = = 0. 75A 4 3. A series RC circuit is connected to voltage source at time t0=0. The relation between the source voltage Vs, the resistance R, the capacitance C, the current i(t) is below: V s = Ri(t) C i(t)dt t 0

16 i(t) i(t) (a) (b) t t i(t) i(t) (c) (d) t t Which one of the following i(t) represents [Gate:04] (a) Given V s = Ri(t) C i(t)dt R 0 t V S i(t) C At t = 0 source Vs is connected to the RC series network. Since there is no charge on capacitor initially so, at t = 0 it acts like a short circuit and the current through the network is i(t) = V s R then as the capacitor starts charging current starts decreasing at the rate the capacitor starts charging

E40M Review - Part 1

E40M Review - Part 1 E40M Review Part 1 Topics in Part 1 (Today): KCL, KVL, Power Devices: V and I sources, R Nodal Analysis. Superposition Devices: Diodes, C, L Time Domain Diode, C, L Circuits Topics in Part 2 (Wed): MOSFETs,